Download Supplementary Exercise 1B Topic 5

Topic 5
Chemical Cell and Electrolysis
Section A
Fill in the blanks
Unit 18
Dry cells in daily life
1
chemical, electrical
2
electrolyte
3
voltmeter, multimeter
4
negative, positive
5
battery
6
service
7
shelf
8
secondary
Unit 19
Simple chemical cells
9
electrons
10
electrochemical
11
zinc, copper, salt bridge
Unit 20
Oxidation and reduction
12
reduction
13
oxidation
14
reducing
15
oxidizing
16
reducing
17
bromide
18
iron(III)
19
chromium(III)
20
manganese(II)
21
sulphate
22
iodine
23
oxidizing agent
24
nitrogen monoxide
25
brown, nitrogen dioxide
26
manganese(IV) oxide
Unit 21
Electrolysis
27
electrolysis
28

electrode
29
electrolytic cell
30
 anode
 cathode
31
 anode
 cathode
32
electrochemical
 concentration
 electrodes
33
electroplating
Section B
True or False
Unit 18
Dry cells in daily life
1
F
The electrolyte in a dry cell is a paste, rather than an aqueous solution.
2
F
In a chemical cell, electrons flow from the negative electrode to the positive electrode in
the external circuit.
3
T
4
F
There are two classes of dry cells: primary cells and secondary cells. Primary cells are not
rechargeable. Secondary cells are rechargeable.
5
F
In a zinc-carbon cell, the carbon rod is the positive electrode. The cylinder casing is made
of zinc, which serves as the negative electrode.
6
F
The electrolyte in an alkaline manganese cell is potassium hydroxide.
7
T
8
F
9
T
10
F
Unit 19
In a silver oxide cell, the negative electrode is zinc powder. The positive electrode is silver
oxide. These materials are separated by an electrolyte containing potassium hydroxide.
A nickel-cadmium cell has cadmium as the negative electrode, nickel(IV) oxide as the
positive electrode.
Simple chemical cells
11
F
In a magnesium-copper cell, electrons flow from the magnesium strip to the copper strip in
the external circuit.
12
F
In the electrochemical series, the position of calcium is higher than that of sodium. The
order is different from that in the reactivity series. This is because calcium atom loses
electrons more readily in cell reactions than in reaction with air, water and dilute acids.
13
F
A salt bridge can be prepared by soaking a piece of filter paper in an ionic salt solution.
14
T
15
T
Unit 20
Oxidation and reduction
16
F
Oxidation is a process in which a substance loses electron(s).
17
F
Not all chemical reactions are redox reactions. For example, neutralization reactions are
not redox reactions.
18
T
19
T
20
F
Dilute nitric acid is an oxidizing agent. It is not used to acidify potassium permanganate
solution because it can react with reducing agents. Usually dilute sulphuric acid is used to
acidify potassium permanganate solution.
21
F
In the reaction between bromine water and sodium sulphite solution, bromine is reduced to
bromide ions while sulphite ions are oxidized to sulphate ions. The resulting reaction
mixture is colourless.
22
F
23
T
24
F
Br2(aq) + 2e–  2Br–(aq)
SO32–(aq) + H2O(l) SO42–(aq) + 2H+(aq) + 2e–
In the reaction between chlorine water and potassium iodide solution, chlorine is reduced
to chloride ions while iodide ions are oxidized to iodine. The resulting reaction mixture is
brown in colour due to the presence of iodine.
Cl2(aq) + 2e– 2Cl–(aq)
2l–(aq)  I2(aq) + 2e–
When acidified potassium permanganate solution acts as an oxidizing agent in a reaction,
the permanganate ions are reduced to manganese(II) ions.
MnO4–(aq) + 8H+(aq) + 5e– Mn2+(aq) + 4H2O(l)
25
T
26
T
27
T
28
T
29
F
30
T
31
F
Magnesium reacts with dilute nitric acid to give nitrogen monoxide.
3Mg(s) + 2NO3–(aq) + 8H+(aq)  3Mg2+(aq) + 2NO(g) + 4H2O(l)
In a zinc-carbon cell, hydrogen is produced and collected on the surface of the positive
electrode. Since hydrogen is a poor conductor of electricity, the accumulation of hydrogen
at the positive electrode may hinder further reactions and decrease the current of the cell.
Manganese(IV) oxide, an oxidizing agent, is used to remove the hydrogen.
Unit 21
Electrolysis
32
F
In an electrolytic cell, reduction takes place at the cathode.
33
F
The type of electrode used for electrolysis would affect the products obtained.
34
F
During the electrolysis of dilute sulphuric acid using platinum electrodes, hydrogen ions
and hydroxide ions are discharged. The net effect is that water is decomposed. The number
of hydrogen ions and sulphate ions from the sulphuric acid remains the same. The
concentration of sulphuric acid increases at the end as water is consumed in the
electrolysis.
35
F
In very dilute sodium chloride solution, there are four kinds of ions:
Cation
Anion
From sodium chloride
From water
Na+ (aq)
H+ (aq)
Cl- (aq)
OH- (aq)
During electrolysis, the sodium ions and hydrogen ions are attracted to the cathode.
A hydrogen ion is a stronger oxidizing agent than a sodium ion. Therefore hydrogen ions
are preferentially discharged (reduced) to form hydrogen gas.
36
F
There are three kinds of ions in acidified water:
Cation
From sulphuric acid
H+ (aq)
From water
H+ (aq)
Anion
SO42- (aq)
OH- (aq)
During electrolysis, the sulphate ions and hydroxide ions are attracted to the anode.
A hydroxide ion is a stronger reducing agent than a sulphate ion. Therefore hydroxide ions
are preferentially discharged (oxidized) to form oxygen gas.
37
T
38
T
39
T
40
T
41
F
In an electroplating process, the object to be plated is made the cathode. The
plating metal is made the anode.
Section C
Multiple Choice
Unit 18
Dry cells in daily life
1
Option B –Electrons flow from the magnesium strip to the copper strip in the external
circuit.
Option C –The mass of the magnesium strip decreases.
Option D –The mass of the copper strip increases.
A
2
C
Option B –Alcohol is a non-conductor.
Option C –A chemical cell consists of two different metals and an electrolyte. The
chemical cell makes the light bulb shine.
3
D
4
A
5
B
6
B
7
A
Option C –The maximum voltage of a zinc-carbon cell is 1.5 V.
Option D –The service life of a zinc-carbon cell is relatively short.
8
C
The voltage of a nickel-cadmium rechargeable cell remains constant over discharge.
9
D
10
A
11
A
A silver oxide cell is not rechargeable.
12
B
Option A –A nickel-cadmium cell can produce a voltage of 1.25 V.
Option C –The electrolyte of a nickel-cadmium cell is potassium hydroxide.
Option D –The voltage of a nickel-cadmium cell remains constant over discharge.
13
B
Type of cell
Zinc-carbon cell
Materials of
electrodes
+ : Carbon
- : Zinc
Material of
electrolyte
Ammonium
chloride
14
C
15
A
16
A
Alkaline
manganese cell
+:
Manganese(IV)
oxide
- : Zinc
Potassium
hydroxide
Silver oxide cell
+ : Silver oxide
- : Zinc
Potassium
hydroxide
Only a small button cell can fit in a hearing aid.
(2)
The electrolyte of the cell is ammonium chloride.
Nickel-cadmium
cell
+ : Nickel(IV)
oxide
- : Cadmium
Potassium
hydroxide
(3)
Zinc is the negative electrode of the cell.
17
D
18
A
(3)
A silver oxide cell can produce a voltage of 1.5 V.
19
A
(3)
A nickel-cadmium cell is able to supply a large current.
20
B
(2)
A zinc-carbon cell is not expensive.
21
A
22
D
23
C
(1)
Using new and used zinc-carbon cells at the same time in an electrical appliance is
not dangerous, but gives poorer results.
24
C
Zinc-carbon dry cells are not rechargeable.
25
A
26
C
The electrolyte of a zinc-carbon cell is a moist paste of ammonium chloride.
Unit 19
Simple chemical cells
27
In the zinc-copper chemical cell, zinc atoms lose electrons and form zinc ions.
A
Zn(s)  Zn2+(aq) + 2e–
The electrons given up by the zinc atoms flow along the conducting wires to the
copper strip. Copper(II) ions near to the copper strip gain these electrons and form
copper atoms.
Cu2+(aq) + 2e– Cu(s)
28
D
Magnesium forms ions more readily than zinc. Magnesium and copper are farther apart in
the electrochemical series than zinc and copper. Therefore the voltage of the cell would
increase if the zinc strip is replaced by a magnesium strip.
29
B
Options A & C –Pure water and alcohol are non-conductors.
Option B –Zinc and copper are farther apart in the electrochemical series than lead and
copper. Therefore the zinc-copper couple gives the highest voltage, making the
light bulb the brightest.
30
D
For the first two chemical cells, metal W is the positive electrode while metals X and Y are
the negative electrodes. Therefore metals X and Y form ions more readily than metal W.
The Y/W couple gives a higher voltage than the X/W couple. Therefore the
difference in the tendency to form ions between metal Y and metal W is greater than that
between metal X and metal W. Hence metal Y forms ions most readily.
For the third chemical cell, the voltmeter gives a negative voltage. Therefore metal
W is the negative electrode while metal Z is the positive electrode. Metal W forms ions
more readily than metal Z.
This gives the descending order of reactivity of the four metals: Y, X, W, Z.
31
D
For a simple chemical cell, the farther apart the two metals are in the electrochemical
series (or the reactivity series), the higher is the voltage of the cell.
32
B
For a simple chemical cell, the closer the two metals are in the electrochemical series (or
the reactivity series), the lower is the voltage of the cell. Copper is closest to silver in the
electrochemical series. Therefore it gives a cell of lowest voltage with silver.
33
C
From the cell of the Mn/Fe couple, we can conclude that Mn forms ions more readily than
Fe.
From the cells of the Fe/Ag couple and Fe/Cu couple, we can conclude that Fe
forms ions more readily than Ag and Cu.
The Fe/Ag couple gives a higher voltage than the Fe/Cu couple. Therefore the
difference in the tendency to form ions between Fe and Ag is greater than that between Fe
and Cu. Hence Ag forms ions least readily.
This gives the descending order of the tendency to form ions for the four metals:
Mn, Fe, Cu, Ag.
34
D
Option
35
B
Metal X
A
Copper
B
Iron
C
Magnesium
D
Silver
Relative tendency to
form ions as
compared to zinc
Zinc forms ions more
readily than copper.
Zinc forms ions more
readily than iron.
Magnesium forms
ions more readily
than zinc.
Zinc forms ions more
readily than silver.
Direction of electron
flow in external circuit
From zinc to copper
From zinc to iron
From magnesium to zinc
From zinc to silver
From the three cells shown, we can conclude than Zn forms ions more readily than Fe
while Mg and Fe form ions more readily than Cu. Therefore the student needs to show the
relative tendency to form ions between Zn and Mg by setting up a cell using the two
metals.
36
A
Options A & D –Zinc forms ions more readily than copper. Electrons flow from the zinc
electrode to the copper electrode in the external circuit. Therefore the
copper electrode is the positive electrode.
Option B –Zinc atoms lose electrons and form zinc ions. Therefore the zinc electrode
decreases in mass.
Option C –Chemical energy is changed into electrical energy.
37
C
Any ionic salt solution that does not react with the reactants of the cell can be used to
prepare a salt bridge.
38
C
(1)
(2)
The electrode made of metal X is the negative electrode.
Metal X forms ions more readily than copper.
39
A
40
A
(3)
The porous pot cannot be replaced by a glass cylinder. The circuit is not complete
as the glass cylinder does not allow ions to move between the two solutions.
41
D
Iron forms ions more readily than copper.
42
C
The position of calcium is higher than that of sodium in the electrochemical series.
Calcium atom loses electrons more readily in cell reactions than in reactions with air,
water and dilute acids.
Unit 20
Oxidation and reduction
43
B
Option A –There is no reaction between copper and dilute sulphuric acid.
Option B –Heating iron in oxygen gives iron(II, III) oxide.
Option C –There is no reaction between zinc and magnesium sulphate solution.
Option D –Adding dilute sodium hydroxide solution to dilute hydrochloric acid is a
neutralization reaction, not a redox reaction.
44
B
Reaction
Heating iron with sulphur
Explanation
0 0
+2-2
Fe + S  FeS
The oxidation number of Fe changes from 0 to +2
while that of S changes from 0 to -2. This is a redox
reaction.
Adding potassium chloride solution to +1-1 +1+5-2
+1-1 +1+5-2
silver nitrate solution
KCl + AgNO3  AgCl + KNO3
This is not a redox reaction.
Burning glucose
Burning glucose gives carbon dioxide and water.
This is a redox reaction.
Heating carbon with lead(II) oxide
Carbon reduces lead(II) oxide to lead. This is a
redox reaction.
45
D
Option A –Electrons flow in the external circuit, not through the salt bridge.
Option B –The magnesium atoms lose electrons to form magnesium ions. Therefore the
mass of the magnesium electrode decreases.
Option C –Reduction occurs at the lead electrode.
Pb2+(aq) + 2e–  Pb(s)
46
C
Permanganate ions MnO4–(aq) are reduced to manganese(II) ions Mn2+(aq).
Chloride ions Cl–(aq) are oxidized to chlorine Cl2(aq).
MnO4–(aq) + 8H+(aq) + 5e–  Mn2+(aq) + 4H2O(l)
2Cl–(aq)  Cl2(aq) + 2e–
47
D
Sulphite ions SO32–(aq) are oxidized to sulphate ions SO42–(aq).
Bromine Br2(aq) is reduced to bromide ions Br–(aq).
SO32–(aq) + H2O(l)  SO42–(aq) + 2H+(aq) + 2e–
Br2(aq) + 2e– 2Br–(aq)
48
A
Iron(III) ions Fe3+(aq) are reduced to iron(II) ions Fe2+(aq).
Iodide ions l–(aq) are oxidized to iodine I2(aq).
Fe3+(aq) + e– Fe2+(aq)
2l–(aq) I2(aq) + 2e–
49
C
50
Option D –Metals above hydrogen in the electrochemical series react with dilute
C
hydrochloric acid to form hydrogen. These metals are stronger reducing agents
than hydrogen. Therefore, they reduce the hydrogen ions in the acid to
hydrogen. The hydrogen ions oxidize the metals to metal ions.
51
B
Chlorine water oxidizes iodide ions to iodine.
52
A
The position of calcium is higher than that of sodium in the electrochemical series.
Therefore calcium is a stronger reducing agent than sodium.
53
D
Element
Atomic number
W
X
Y
Z
Name of
element
Neon
Magnesium
Silicon
Chlorine
10
12
14
17
Chlorine is an oxidizing agent.
54
C
Suppose the oxidation number of S in H2SO3 is x.
(+1) x 2 + x + (–2) x 3
x
= +4
55
D
56
A
57
C
=0
Suppose the oxidation number of Cl in HClO4 is x.
(+1) + x + (–2) x 4 = 0
x
= +7
–2
FeS
+6
(NH4)2SO4
Option
A
B
C
D
+4
PbO2
+3
Fe2O3
+4
NO2
-2
OH-
+4
SO2
+2
Na2S2O3
+7
MnO4+4
CO2
+4
CO3 2+6
Cr2O72-
58
B
0
-3
Stage I : N2  NH3
-3
+2
Stage II : NH3  NO
+2
+4
Stage III: NO  NO2
+4
+5
Stage IV : NO2  HNO3
Change in oxidation number
3
5
2
1
59
D
60
A
+5
–1
ClO3–  Cl–
–3
+6
0
+3
(NH4)2Cr2O7  N2 + 4H2O + Cr2O3
61
A
Element X is calcium.
62
C
Option
A
B
C
D
63
Equation
+3
+2
Fe3+(aq)  Fe2+(aq)
+5
+2
NO3-(aq)  NO(g)
+7
+2
MnO4-(aq)  Mn2+(aq)
+6
+4
H2SO4(aq)  SO2(g)
Change in oxidation number of the underlined element
1
3
5
2
D
Option
A
B
C
D
Equation
KOH + KNO3  KNO3 + H2O
0
-1
8NH3 + 3Cl2  6NH4Cl + N2
+2
0
Zn + FeSO4  ZnSO4 + Fe
-1
0
2HBr + H2SO4  SO2 + 2H2O + Br2
Explanation
This is not a redox reaction.
Cl2 becomes reduced.
Iron(II) ions are reduced.
Bromide ions are oxidized to
bromine.
64
D
Option
A
Equation
+2
+3
2FeCl2 + Cl2  2FeCl3
Explanation
Iron(II) ions are oxidized to iron(III) ions.
B
This is not a redox reaction.
MgCO3 + 2HCl  MgCl2 + CO2 +H2O
-1
0
Fe2(SO4)3 + 2Kl  2FeSO4 + K2SO4 + I2
+2
0
2NH3 + 3CuO  3Cu + N2 + 3H2O
C
D
65
Equation
C + 2PbO  CO2 + 2Pb
B
Br2 + 2Kl  2KBr + I2
C
Zn + CuSO4  ZnSO4 + Cu
D
2Na + Cl2  2NaCl
Explanation
Carbon reduces lead(II) oxide to lead. It is a reducing
agent.
Bromine oxidizes iodide ions to iodine. It is an oxidizing
agent.
Copper(II) sulphate oxidizes zinc to zinc ions. It is an
oxidizing agent.
Chlorine oxidizes sodium to sodium ions. It is an
oxidizing agent.
B
Option
A
B
C
D
67
Copper(II) oxide is reduced.
A
Option
A
66
Iodide ions are oxidized to iodine.
A
Equation
0
+2
+2
0
Mg(s)+ Pb2+(aq)  Mg2+(aq) + Pb(s)
+6 -2
+1
+6 -2
+1 -2
2CrO42-(aq) + 2H+(aq)  Cr2O72-(aq) + H2O(l)
+5-2
-1
0
2KClO3(s)  2KCl(s) + 3O2(g)
+2
0
0
+4
2CuO(s) + C(s)  2Cu(s) + CO2(g)
Explanation
This is a redox reaction.
This is not a redox reaction.
This is a redox reaction.
This is a redox reaction.
In the reaction, acidified potassium permanganate solution oxidizes the iodide ions to
iodine.
MnO4–(aq) + 8H+(aq) + 5e–  Mn2+(aq) + 4H2O(l)
2l–(aq)  I2(aq) + 2e–
Option A –The reaction mixture is brown in colour due to the presence of iodine.
Option B –Iodide ions are oxidized.
Option C –The acidified potassium permanganate solution is the oxidizing agent.
Option D –The oxidation number of manganese changes from +7 to +2.
+7
+2
MnO4–(aq)  Mn2+(aq)
68
D
In the reaction, acidified potassium dichromate solution oxidizes the sulphite ions to
sulphate ions.
Cr2O72–(aq) + 14H+(aq) + 6e– 2Cr3+(aq) + 7H2O(l)
SO32–(aq) + H2O(l)  SO42–(aq) + 2H+(aq) + 2e–
Option A –The colour of the acidified potassium dichromate solution changes to green due
to the formation of chromium(III) ions.
Option D –The oxidation number of chromium changes from +6 to +3.
+6
+3
2–
Cr2O7 (aq)  Cr3+(aq)
69
B
In the reaction, acidified potassium permanganate solution oxidizes the iron(II) ions to
iron(III) ions.
MnO4-(aq) + 8H+(aq) + 5e-  Mn2+(aq) + 4H2O(l)
Fe2+(aq)  Fe3+(aq) + eOption A –Permanganate ions are reduced to manganese(II) ions.
Option B –Iron(II) sulphate solution is the reducing agent. Iron(II) ions are oxidized to
iron(III) ions.
Option C –The oxidation number of potassium remains the same.
Option D –The reaction mixture appears yellow due to the presence of iron(III) ions.
70
A
In the reaction, chlorine water oxidizes the bromide ions to bromine (which is brown in
colour).
Cl2(aq) + 2e-  2Cl-(aq)
2Br-(aq)  Br2(aq) + 2eOption B –Chlorine water acts as an oxidizing agent.
Option C –Bromide ions are oxidized.
Option D –The oxidation number of chlorine changes from 0 to –.
0
-1
Cl2(aq)  2Cl-(aq)
71
C
In the reaction between copper and dilute nitric acid, nitrate ions are reduced to nitrogen
monoxide.
Nitrogen monoxide gas is colourless. When this gas mixes with air, it reacts with
oxygen in the air to give brown nitrogen dioxide gas.
2NO(g) + O2(g)  2NO2(g)
colourless
brown
72
B
In the reaction between copper and concentrated nitric acid, nitrate ions are reduced to
nitrogen dioxide which is brown in colour.
Option C –Nitrogen dioxide gas is toxic.
73
B
74
A
75
A
In the reaction, the nitrate ion in nitric acid is reduced to nitrogen monoxide.
+5
+2
+
NO3 (aq) + 4H (aq) + 3e  NO(g) + 2H2O(l)
Option B –Concentrated nitric acid reacts with magnesium to give nitrogen dioxide.
Mg(s) + 2NO3-(aq) + 4H+(aq)  Mg2+(aq) + 2NO2(g) + 2H2O(l)
Option C –Concentrated nitric acid reacts with copper to give nitrogen dioxide.
Cu(s) + 2NO3-(aq) + 4H+(aq)  Cu2+(aq) + 2NO2(g) + 2H2O(l)
Option D –The corrosive nature of concentrated nitric acid is due to its powerful oxidizing
property.
76
C
Copper does not react with dilute ethanoic acid, dilute hydrochloric acid and dilute
sulphuric acid.
77
B
We can represent the reaction between copper in the alloy and concentrated nitric acid by
the following equation:
Cu(s) + 2NO3-(aq) + 4H+(aq)  Cu2+(aq) + 2NO2(g) + 2H2O(l)
We can represent the reaction between zinc in the alloy and concentrated nitric acid
by the following equation:
Zn(s) + 2NO3-(aq) + 4H+(aq)  Zn2+(aq) + 2NO2(g) + 2H2O(l)
The resulting solution contains Cu2+(aq)and Zn2+(aq). It is pale blue in colour.
78
A
79
D
(1)
(2)
80
B
81
D
(1)
(2)
In the left beaker, the solution changes from green to yellow gradually. This is
because iron(II) ions Fe2+ (aq) lose electrons and change to iron(III) ions Fe3+(aq).
Fe2+(aq)  Fe3+(aq) + eIn the right beaker, the purple colour of the solution fades out gradually. This is
because the permanganate ions MnO4-(aq)gain electrons and change to
manganese(II) ions Mn2+(aq).
MnO4-(aq) + 8H+(aq) + 5e-  Mn2+(aq) + 4H2O(l)
In the left beaker, the potassium iodide solution changes from colourless to brown.
This is because iodide ions I- (aq) lose electrons and change to iodine I2(aq).
2I-(aq)  I2(aq) + 2eIn the right beaker, the solution changes from yellow to green gradually. This is
because iron(III) ions Fe3+(aq) gain electrons and change to iron(II) ions Fe2+(aq).
Fe3+(aq) + e-  Fe2+(aq)
82
A
Magnesium is above hydrogen in the electrochemical series. It is a stronger reducing agent
than hydrogen. Therefore it can reduce the hydrogen ions in dilute hydrochloric acid to
hydrogen. The hydrogen ion oxidizes magnesium to magnesium ions.
Mg(s) + 2H+(aq)  Mg2+(aq) + H2(g)
(3)
The oxidation number of hydrogen changes from +1 to 0.
83
D
In the reaction, magnesium displaces copper from the copper(II) sulphate solution.
Mg(s) + Cu2+(aq)  Mg2+(aq) + Cu(s)
(1)
The oxidation number of magnesium changes from 0 to +2. It is oxidized.
(2)
The oxidation number of Cu2+ changes from +2 to 0. Copper(II) sulphate solution
acts as an oxidizing agent. It is reduced.
(3)
The colour of the copper(II) sulphate solution fades out because the concentration
of copper(II) ions decreases.
84
A
(2)
(3)
85
D
(1)
(2)
The oxidation number of oxygen is – in peroxides, such as sodium peroxide,
Na2O2.
The oxidation number of alkaline earth metals is +2 in all their compounds.
Magnesium displaces iron from iron(II) sulphate solution.
Mg(s) + FeSO4(aq)  MgSO4(aq) + Fe(s)
The oxidation number of iron changes from +2 to 0.
Iron reacts with dilute hydrochloric acid to give iron(II) chloride and hydrogen.
Fe(s) + 2HCl(aq)  FeCl2(aq) + H2(g)
(3)
The oxidation number of iron changes from 0 to +2.
Carbon can extract iron from iron(III) oxide. The oxidation number of iron changes
from +3 to 0.
86
C
In the reaction, bromine water oxidizes the sulphite ions to sulphate ions.
Br2(aq) + 2e-  2Br-(aq)
SO32-(aq) + H2O(l)  SO42-(aq) + 2H+(aq) + 2e(1)
The sulphite ions are oxidized.
(2)
The brown bromine water becomes colourless due to the formation of bromide ions.
87
C
In the reaction, iron(III) ions oxidize the iodide ions to iodine.
Fe3+(aq) + e-  Fe2+(aq)
2I-(aq)  I2(aq) + 2e(1)
Potassium iodide solution acts as a reducing agent.
(2)
Iron(III) ions are reduced to iron(II) ions.
(3)
The resulting reaction mixture is brown in colour due to the presence of iodine.
88
B
(1)
(3)
Na2SO3(aq) can reduce permanganate ions to manganese(II) ions. The sulphite
ions are oxidized to sulphate ions.
MnO4-(aq) + 8H+(aq) + 5e-  Mn2+(aq) + 4H2O(l)
SO32-(aq) + H2O(l)  SO42-(aq) + 2H+(aq) + 2eFeSO4(aq) can reduce permanganate ions to manganese(II) ions. The iron(II) ions
are oxidized to iron(III) ions.
MnO4-(aq) + 8H+(aq) + 5e-  Mn2+(aq) + 4H2O(l)
Fe2+(aq)  Fe3+(aq) + e-
89
C
Equation
+2
-2 +1
+2 -2 +1
2+
(1) Cu + 2OH-  Cu(OH)2
+4
0
+6
-1
22(2) SO3 + Cl2 + H2O  SO4 + 2Cl- + 2H+
+1
0
+1
0
(3) 2H2O + 2Na  2NaOH + H2
90
Explanation
This is not a redox reaction.
The oxidation number of S changes from +4
to +6 while that of Cl changes from 0 to -1.
The oxidation number of Na changes from 0
to +1 while that of H changes from +1 to 0.
D
Equation
-3+1 -2 +1 -3 +1 +1 -2
(1) NH4+ + OH-  NH3 + H2O
0
-3
0
-1
(2) 3Cl2 + 8NH3  N2 + 6NH4Cl
0
+6
+2
+4
(3) Cu + 2H2SO4  CuSO4 + SO2 + 2H2O
Explanation
This is not a redox reaction.
The oxidation number of N changes from -3 to 0
while that of Cl changes from 0 to -1.
The oxidation number of Cu changes from 0 to +2
while that of S changes from +6 to +4.
91
A
An atom of the element X has an electronic arrangement of 2,8,8,1. X is potassium.
(3)
X burns with a purple flame.
92
C
(1)
(2)
In beaker A, the colour of the bromine water fades out gradually because bromine
water is converted to bromide ions (which is colourless).
Br2(aq) + 2e-  2Br-(aq)
In beaker B, sulphite ions are oxidized to sulphate ions.
SO32-(aq) + H2O(l)  SO42-(aq) + 2H+(aq) + 2e-
93
C
(1)
The solution formed contains copper(II) ions and is blue in colour.
94
B
(2)
Very dilute nitric acid acts like a typical acid. It does not react with unreactive
metals, such as copper.
95
A
(2)
(3)
The oxidation number of zinc changes from 0 to +2. Zinc is oxidized.
The oxidation number of hydrogen remains the same (+1).
96
B
(1)
(2)
The nitrate ion in concentrated nitric acid is reduced to nitrogen dioxide, a brown
gas.
+5
+4
+
NO3 (aq) + 2H (aq) + e  NO2(g) + H2O(l)
The oxidation number of nitrogen changes from +5 to +4.
(1)
(2)
Litmus solution gives a red colour in both acids.
Dilute hydrochloric acid gives a white precipitate with silver nitrate solution while
(3)
dilute nitric acid does not.
Ag+(aq) + Cl-(aq)  AgCl(s)
Sodium carbonate reacts with both acids to give carbon dioxide gas and a
colourless solution.
CO32-(aq) + 2H+(aq)  H2O(l) + CO2(g)
97
98
B
A
(1)
Reduction occurs at the positive electrode:
2NH4+(aq) + 2e-  2NH3(aq) + H2(g)
Hydrogen is produced and collected on the surface of the positive electrode. Since
(2)
(3)
99
B
100
C
101
A
102
A
hydrogen is a poor conductor of electricity, the accumulation of hydrogen at the
positive electrode may hinder further reactions and decrease the current of the cell.
Manganese(IV) oxide, an oxidizing agent, is used to remove the hydrogen.
2MnO2(s) + H2(g)  Mn2O3(s) + H2O(l)
Ammonium chloride is the electrolyte.
There is a slow direct reaction between the zinc electrode and ammonium ions. The
zinc case becomes thinner upon storage.
Sodium is a stronger reducing agent than magnesium.
The reaction between silver nitrate solution and dilute hydrochloric acid gives silver
chloride precipitate.
Unit 21
Electrolysis
103
D
Option A –Electrolysis is a process in which electrical energy is converted into chemical
energy.
Option B –Oxidation occurs at the anode.
Option C –The electrode connected to the positive terminal of the power supply is the
anode.
104
B
There are three kinds of ions in acidified water:
Cation
From sulphuric acid
From water
+
H (aq)
H+(aq)
Anion
2-
SO4 (aq)
OH-(aq)
The sulphate ions and hydroxide ions are attracted to the anode. The hydrogen ions
are attracted to the cathode.
At the anode
A hydroxide ion is a stronger reducing agent than a sulphate ion. Therefore hydroxide ions
are preferentially discharged (oxidized) to form oxygen gas.
At the cathode
Hydrogen ions are discharged (reduced) to form hydrogen gas.
105
B
Option A –When hydrogen ions and hydroxide ions are discharged, more water molecules
ionize. The net effect is that water is decomposed. The number of hydrogen
ions and sulphate ions from the sulphuric acid remains the same. The
concentration of sulphuric acid increases at the end as water is consumed in the
electrolysis.
Option C –The gas collected above electrode X is oxygen. It can relight a glowing splint.
Option D –The gas collected above electrode Y is hydrogen. It burns with a 厜 op–
sound.
106
C
In the solution, there are three kinds of ions:
Cation
Anion
From potassium hydroxide
K+(aq)
OH-(aq)
From water
H+(aq)
OH-(aq)
At the anode
Hydroxide ions are discharged (oxidized) to form oxygen gas.
At the cathode
A hydrogen ion is a stronger oxidizing agent than a potassium ion. Therefore hydrogen
ions are preferentially discharged (reduced) to form hydrogen gas.
Overall cell reaction
Volume of oxygen: volume of hydrogen is 1:2. This can be shown by the equation for the
overall reaction:
2H2O(l)  2H2(g) + O2(g)
107
B
In the dilute sodium nitrate solution, there are four kinds of ions:
Cation
Anion
From sodium nitrate
Na+(aq)
NO3-(aq)
From water
H+(aq)
OH-(aq)
At the anode
A hydroxide ion is a stronger reducing agent than a nitrate ion. Therefore hydroxide ions
are preferentially discharged (oxidized) to form oxygen gas.
At the cathode
A hydrogen ion is a stronger oxidizing agent than a sodium ion. Therefore hydrogen ions
are preferentially discharged (reduced) to form hydrogen gas.
108
C
Water ionizes continuously to replace the hydroxide ions discharged at the anode
(electrode X).
H2O(l)
H+(aq) + OH- (aq)
Thus there is an excess of hydrogen ions near the anode (electrode X). The solution
there becomes acidic.
At the same time, water ionizes continuously to replace the hydrogen ions
discharged at the cathode (electrode Y). Thus there is an excess of hydroxide ions near the
cathode (electrode Y). The solution there becomes alkaline.
The solution near electrode X will turn red while the solution near electrode Y will
turn blue.
109
A
The concentration of sodium nitrate solution increases as water is consumed in the
electrolysis.
110
A
Cations at lower positions of the electrochemical series are more readily discharged
because they are stronger oxidizing agents (i.e. they gain electrons more readily).
Therefore Ag+ (aq) is most readily discharged.
111
B
Anions at higher positions of the electrochemical series are more readily discharged
because they are stronger reducing agents (i.e. they lose electrons more readily). Therefore
OH- (aq) is most readily discharged.
112
D
Cations at lower positions of the electrochemical series are more readily discharged
because they are stronger oxidizing agents (i.e. they gain electrons more readily.) The
position of H+ (aq) in the electrochemical series is lower than that of K+(aq). Therefore
H+(aq) accepts an electron more readily.
113
D
In dilute copper(II) sulphate solution, there are four kinds of ions:
Cation
Anion
From copper(II) sulphate
Cu2+(aq)
SO42-(aq)
From water
H+(aq)
OH-(aq)
The copper(II) ions and hydrogen ions are attracted to the cathode.
At the cathode
A copper(II) ion is a stronger oxidizing agent than a hydrogen ion. Hence, copper(II) ions
are preferentially discharged (reduced) to form a deposit of copper on the cathode.
114
C
Option A –The blue colour of the solution fades out because the concentration of copper(II)
ions decreases.
Option B –Copper(II) ions and hydroxide ions are consumed in the electrolysis. Hydrogen
ions and sulphate ions remain in the solution. Thus the solution eventually
becomes sulphuric acid.
115
B
In dilute copper(II) chloride solution, there are four kinds of ions:
Cation
Anion
From copper(II) chloride
Cu2+(aq)
Cl-(aq)
From water
H+(aq)
OH-(aq)
The chloride ions and hydroxide ions are attracted to the anode. The copper(II) ions and
hydrogen ions are attracted to the cathode.
At the anode
Copper is a stronger reducing agent than hydroxide and chloride ions. The copper anode
dissolves to form copper(II) ions (oxidized).
At the cathode
A copper(II) ion is a stronger oxidizing agent than a hydrogen ion. Therefore copper(II)
ions are preferentially discharged (reduced) to form a deposit of copper on the cathode.
116
C
The net effect is the transfer of copper from the anode to the cathode. The rate at which
copper deposits on the cathode is equal to the rate at which the copper anode dissolves.
Increase in mass of cathode = decrease in mass of anode
The concentration of copper(II) ions in the electrolyte remains the same.
117
A
Overall cell reaction
Cu(s)  Cu(s)
(anode)
(cathode)
Changes in the solution
The net effect is the transfer of copper from the anode to the cathode. The rate at which
copper deposits on the cathode is equal to the rate at which the copper anode dissolves.
Increase in mass of cathode = decrease in mass of anode
The solution remains unchanged.
118
D
In dilute copper(II) sulphate solution, there are four kinds of ions:
Cation
Anion
From copper(II) sulphate
Cu2+(aq)
SO4-(aq)
From water
H+(aq)
OH-(aq)
The sulphate ions and hydroxide ions are attracted to the anode. The copper(II)
ions and hydrogen ions are attracted to the cathode.
At the anode
A hydroxide ion is a stronger reducing agent than a sulphate ion. Therefore hydroxide ions
are preferentially discharged (oxidized) to form oxygen gas.
At the cathode
A copper(II) ion is a stronger oxidizing agent than a hydrogen ion. Hence, copper(II) ions
are preferentially discharged (reduced) to form a deposit of copper on the cathode.
Changes in the solution
Copper(II) ions and hydroxide ions are consumed in the electrolysis. Hydrogen ions and
sulphate ions remain in the solution. Thus the solution eventually becomes sulphuric acid.
119
B
In concentrated sodium chloride solution, there are four kinds of ions:
Cation
Anion
From sodium chloride
Na+(aq)
Cl-(aq)
From water
H+(aq)
OH-(aq)
At the anode
The concentration of chloride ions in the solution is much greater than that of hydroxide
ions. Chloride ions are preferentially discharged (oxidized) to form chlorine gas.
2Cl-(aq)  Cl2(g) + 2eAt the cathode
A sodium ion is a weaker oxidizing agent than a hydrogen ion. However, if mercury is
used as the cathode, sodium ions are preferentially discharged (reduced) to form sodium
metal. The metal formed dissolves in the mercury cathode to form an alloy called sodium
amalgam.
Na+(aq) + e- + Hg(l)  Na/Hg(l)
sodium amalgam
The sodium amalgam then moves towards the water. The sodium reacts with water
to form sodium hydroxide and hydrogen.
2Na/Hg(l) + 2H2O(l)  2NaOH(aq) + H2(g) + 2Hg(l)
120
B
Sodium ions and chloride ions are consumed in the electrolysis. Thus the sodium chloride
solution becomes more and more dilute.
121
A
Option A –The mass of electrode B increases due to copper deposits. Therefore electrodes
A and C are anodes while electrodes B and D are cathodes. X is the positive
terminal of the battery.
Option B –Electrode A is the anode.
Option C –Reduction occurs at electrode B (cathode).
Option D –The net effect is the transfer of copper from electrode A to electrode B. The
rate at which copper deposits on the cathode is equal to the rate at which the
copper anode dissolves. The concentration of copper(II) ions in the electrolyte
remains the same. The blue colour of the copper(II) sulphate solution does not
change.
122
C
In dilute sodium iodide solution, there are four kinds of ions:
Cation
Anion
From sodium iodide
Na+(aq)
I-(aq)
From water
H+(aq)
OH-(aq)
The iodide ions and hydroxide ions are attracted to the anode. The sodium ions
and hydrogen ions are attracted to the cathode.
At the anode
The concentration of iodide ions in the solution is much greater than that of hydroxide ions.
Therefore iodide ions are preferentially discharged (oxidized) to form iodine.
123
C
During the electrolysis of dilute sodium bromide solution using carbon electrodes,
hydrogen ions and bromide ions are consumed, sodium ions and hydroxide ions remain in
the solution. The solution eventually becomes sodium hydroxide solution.
124
D
125
D
126
C
127
A
In the electrolysis of dilute copper(II) sulphate solution using copper electrodes, the copper
anode dissolves to form copper(II) ions. Copper(II) ions are preferentially discharged to
form a deposit of copper on the cathode. The net effect is the transfer of copper from the
anode to the cathode. The concentration of the copper(II) sulphate solution remains the
same.
The following figure shows the refining process of copper.
Options A & B –During the refining process, the silver impurity in the impure copper rod
settles at the bottom of the container. The mass of the impure copper rod
decreases.
Option C –At the pure copper rod, copper(II) ions are discharged to form copper.
Reduction takes place at the pure copper rod.
Option D –The process makes the copper more pure.
128
B
(2)
A d.c. power supply is used in electrolysis.
129
D
(3)
During the refining of impure copper containing zinc, the impure copper is the
anode. Zinc forms ions more readily than copper. Therefore zinc in the anode gives
up electrons first.
Zn(s)  Zn2+(aq) + 2e-
130
C
(1) & (3)
Reduction occurs at the cathode.
131
B
Which ions are preferentially discharged during electrolysis depends on a number of
factors, including:
‧the position of ions in the electrochemical series;
‧the concentration of ions in the solution; and
‧the nature of the electrodes.
132
A
(1)
In concentrated sodium chloride solution, there are four kinds of ions:
From sodium chloride
Cation
Anion
Na+(aq)
Cl-(aq)
From water
H+(aq)
OH-(aq)
The chloride ions and hydroxide ions are attracted to the anode.
At the anode
The concentration of chloride ions in the solution is much greater than that of
hydroxide ions. Therefore chloride ions are preferentially discharged (oxidized) to
form chlorine gas.
(2)
At the cathode
A hydrogen ion is a stronger oxidizing agent than a sodium ion. Therefore
hydrogen ions are preferentially discharged (reduced) to form hydrogen gas.
Water ionizes continuously to replace the hydrogen ions discharged at the
cathode. Thus there is an excess of hydroxide ions near the cathode.
(3)
133
Hydrogen ions and chloride ions are consumed in the electrolysis. Sodium ions and
hydroxide ions remain in the solution. Eventually, the solution becomes sodium
hydroxide solution. The phenolphthalein turns red.
A
Substabce
Potassium sulphate solution
Silver nitrate solution
Molten sodium chloride
134
B
Anode
Oxygen
Oxygen
Sodium
Products at
Cathode
Hydrogen
Silver
chlorine
In concentrated sodium iodide solution, there are four kinds of ions:
Cation
Anion
+
From sodium iodide
Na (aq)
I-(aq)
From water
H+(aq)
OH-(aq)
The iodide ions and hydroxide ions are attracted to the anode. The sodium ions and
hydrogen ions are attracted to the cathode.
(1)
(2)
(3)
135
At the anode
The concentration of iodide ions in the solution is much greater than that of
hydroxide ions. Iodide ions are preferentially discharged (oxidized) to form iodine.
At the cathode
A hydrogen ion is a stronger oxidizing agent than a sodium ion. Therefore
hydrogen ions are preferentially discharged (reduced) to form hydrogen gas.
Hydrogen ions and iodide ions are consumed in the electrolysis. Sodium ions and
hydroxide ions remain in the solution. Eventually, the solution becomes sodium
hydroxide solution.
A
Electrolyte
Dilte copper(II) chloride solution
Molten lead(II) bromide
Dilute sodium hydroxide soluition
136
D
(1)
(2)
(3)
137
A
138
C
Products of electrolysis
Chlorine gas and copper
Lead and bromine gas
Hydrogen gas and oxygen gas
A copper(II) ion is a stronger oxidizing agent than a hydrogen ion. Hence,
copper(II) ions are preferentially discharged (reduced) to form a deposit of copper
on the cathode.
A silver ion is a stronger oxidizing agent than a hydrogen ion. Hence, silver ions
are preferentially discharged (reduced) to form a deposit of silver on the cathode.
A sodium ion is a weaker oxidizing agent than a hydrogen ion. However, if
mercury is used as the cathode, sodium ions are preferentially discharged (reduced)
to form sodium metal.
Electrolyte
Concentrate sodium chloride solution
Dilute magnesium sulphate solution
Dilute copper(II) sulphate solution
Product at anode
Chlorine
Oxygen
oxygen
139
B
(2)
Iron is extracted by carbon reduction.
140
D
During the refining process, the copper is gradually transferred from the anode to the
cathode.
141
D
142
D
143
A
144
B
Reasons for using carbon anodes include:
‧carbon is cheap;
‧carbon is inert;
‧carbon is a good conductor of electricity.
145
D
During the electrolysis of concentrated sodium bromide solution using carbon electrodes,
bromine is produced at the anode. Bromine, rather than oxygen, is produced due to the
concentration effect.
146
A
147
D
Carbon anodes are used in the electrolysis of brine.
Chlorine gas can attack platinum.
Topic 5 Miscellaneous
148
D
149
A
The positive electrode of a nickel-cadmium cell is nickel(IV) oxide.
150
C
Magnesium and silver are farthest apart in the electrochemical series. The magnesium-
silver couple gives the highest voltage.
151
D
Aluminium and copper are farthest apart in the electrochemical series. The aluminiumcopper couple gives the highest voltage.
152
B
Electrons flow from metals P, Q and S to copper when tested. Therefore metals P, Q and S
form ions more readily than copper.
The metal Q-copper couple gives the highest voltage while metal P-copper
gives the lowest voltage. Therefore the difference in the tendency to form ions between
metal Q and copper is the greatest while that between metal P and copper is the smallest.
Therefore the descending order of tendency to form ions of the three metals are: Q
> S > P.
Electrons flow from copper to metal R. Therefore copper forms ions more readily
than metal R. Metal R has the least tendency to form ions among the four metals.
153
C
154
B
In the chemical cell, zinc atoms lose electrons and form zinc ions. The electrons given up
by the zinc atoms flow along the conducting wires to electrode Y. Copper(II) ions near to
electrode Y gain these electrons and form copper atoms.
Element
W
X
Y
Z
155
C
Number of protons in atom
6
8
11
18
Name of element
Carbon
Oxygen
Sodium
Argon
Option A –Metal X is the negative electrode.
Option B –Metal Y is the positive electrode.
Option D –A salt bridge is simply a piece of filter paper soaked in an ionic salt solution.
156
B
From the electrochemical cell formed by connecting half-cells A and C, we can conclude
that Y forms ions more readily than Cu. From the electrochemical cell formed by
connecting half-cells B and C, we can conclude that X forms ions more readily than Y.
This gives the descending order of the tendency to form ions for the three metals:
X, Y, Cu.
157
C
X is more reactive than Y and thus it can displace Y from a solution of a compound of Y.
158
A
Sodium reacts with water to give sodium hydroxide and hydrogen.
0
+1
+1
0
2Na(s) + 2H2O(l)  2NaOH(aq) + H2(g)
The oxidation number of sodium changes from 0 to +1 while that of hydrogen changes
from +1 to 0.
159
B
Suppose the oxidation number of Cl in KClO3 is x.
(+1) + x + (-2) x 3 = 0
x = +5
160
C
Suppose the oxidation number of Al in [Al(OH)4]–is x.
x + (-1) x 4 = -1
x = +3
161
A
162
B
–3
NH3
+2
NO
+4
NO2
Option
A
B
C
D
163
C
+5
HNO3
+4
-2
SO2 K2S
+6
+6
SO3 H2SO4
-2
+4
H2S Na2SO3
+2
+6
Na2S2O3 K2SO4
(NH4)2Cr2O7 consists of NH4+ ions and Cr2O72-ions.
Suppose the oxidation number of Cr in Cr2O72–is x.
2x + (-2) x 7 = -2
x = +6
164
A
Conversion
+5
+4
NO3- NO2
+4
+6
SO32-  SO42+5
+2
NO3-  NO
Change in oxidation number of the underlined element
1
2
3
+7
+2
MnO4-  Mn2+
5
165
A
In the reaction, zinc displaces iron from the iron(II) sulphate solution.
Zn(s) + Fe2+(aq)  Zn2+(aq) + Fe(s)
Option B –The oxidation number of zinc changes from 0 to +2.
Option C –The oxidation number of iron changes from +2 to 0.
Option D –The green colour of the iron(II) sulphate solution fades out because the
concentration of iron(II) ions decreases.
166
C
In the reaction, chlorine water oxidizes bromide ions to bromine (which is brown in
colour).
167
A
+3
0
+5 –1
PCl3 + Cl2  PCl5
The oxidation number of phosphorus changes from +3 to +5 while that of chlorine changes
from 0 to –.
168
D
+3
–2
+2
0
Fe2(SO4)3 + H2S  2FeSO4 + S + H2SO4
The oxidation number of iron changes from +3 to +2 while that of sulphur changes from –
to 0.
169
C
Equation
+2 +6
+3
+6
+4
2FeSO4 +  Fe2O3 + SO3 + SO2
0 +1
+3
0
2Al + 6HCl  2AlCl3 + 3H2
+1+1+4-2
+1 +4-2 +1 -2 +4-2
2NaHCO3 +  Na2CO3 + H2O + CO2
+3
+2
0 +4
Fe2O3 + 3CO  + Fe + 3CO2
170
Explanation
The oxidation number of Fe changes from +2 to
+3 while that of S changes from +6 to +4.
The oxidation number of Al changes from 0 to +3
while that of H changes from +1 to 0.
This is not a redox reaction.
The oxidation number of Fe changes from +3 to 0
while that of C changes from +2 to +4.
B
Equation
2HCl + MgO  MgCl2 + H2O
+4
0
SO2 + 2H2S  3S + 2H2O
-1
0
Explanation
This is not a redox reaction.
SO2 undergoes reduction.
KBr undergoes oxidation.
2KBr + Cl2  2KCl + Br2
2NH3 + 3CuO  3Cu + N2 + 3H2O
171
NH3 undergoes oxidation.
A
Reaction
4H2 + Fe3O4  3Fe + 4H2O
SO2 + 2Mg  2MgO + S
Pb(NO3)2 + H2SO4  PbSO4 + 2HNO3
Zn + 2AgNO3  Zn(NO3)2 + 2Ag
Explanation
Hydrogen reduces iron(II, III) oxide to iron. It acts
as a reducing agent.
Magnesium reduces sulphur dioxide to sulphur. It
acts as a reducing agent.
This is not a redox reaction.
Zinc reduces silver ions to silver. It acts as a
reducing agent.
172
B
Iron(II) sulphate solution reacts with ammonia solution to give a green precipitate, iron(II)
hydroxide.
Fe2+(aq) + 2OH-(aq)  Fe(OH)2(s)
173
D
In the reaction, acidified potassium permanganate solution oxidizes iron(II) ions to iron(III)
ions.
MnO4-(aq) + 8H+(aq) + 5e-  Mn2+(aq) + 4H2O(l)
Fe2+(aq)  Fe3+(aq) + eOption A –The acidified potassium permanganate solution acts as an oxidizing agent.
Option B –The resulting solution is yellow in colour due to the presence of iron(III) ions.
Option C –The oxidation number of manganese changes from +7 to +2.
174
D
Sodium sulphate solution is not a reducing agent.
175
A
A solution containing Ca2+(aq) gives a white precipitate with dilute sodium hydroxide
solution.
Ca2+(aq) + OH-(aq)  Ca(OH)2(s)
Chlorine water oxidizes bromide ions to bromine (which is brown in colour).
Cl2(aq) + 2Br-(aq)  2Cl-(aq) + Br2(aq)
176
D
Equation
Fe + S  FeS
H2SO4 + CuO  CuSO4 + H2O
2FeCl2 + Cl2  2FeCl3
ZnSO4 + Mg  Zn + MgSO4
Explanation
S becomes reduced,
This is not a redox reaction.
Cl2 becomes reduced.
Mg becomes oxidized.
177
A
Option A –Adding magnesium to dilute nitric acid gives nitrogen monoxide.
Option C –Sodium reacts with the water in dilute copper(II) sulphate solution to give
hydrogen.
178
D
In dilute sodium nitrate solution, there are four kinds of ions:
Cation
Anion
From sodium nitrate
Na+(aq)
NO3-(aq)
From water
H+(aq)
OH-(aq)
The nitrate ions and hydroxide ions are attracted to the anode. The sodium ions and
hydrogen ions are attracted to the cathode.
Option A –At the anode (electrode X)
A hydroxide ion is a stronger reducing agent than a nitrate ion. Therefore
hydroxide ions are preferentially discharged (oxidized) to form oxygen gas.
Option B –At the cathode (electrode Y)
A hydrogen ion is a stronger oxidizing agent than a sodium ion. Therefore
hydrogen ions are discharged (reduced) to form hydrogen gas.
Option C –Water ionizes continuously to replace the hydroxide ions discharged at the
anode (electrode X). Thus there is an excess of hydrogen ions near the electrode
X and the solution there becomes acidic. The universal indicator turns red after
some time.
Option D –The sodium nitrate solution becomes more concentrated as water is
decomposed in the process.
179
C
In concentrated potassium bromide solution, there are four kinds of ions:
Cation
Anion
From potassium bromide
K+(aq)
Br-(aq)
From water
H+(aq)
OH-(aq)
The bromide ions and hydroxide ions are attracted to the anode. The potassium
ions and hydrogen ions are attracted to the cathode.
At the anode
The concentration of bromide ions in the solution is much greater than that of hydroxide
ions. Therefore bromide ions are preferentially discharged (oxidized) to form bromine.
At the cathode
A hydrogen ion is a stronger oxidizing agent than a potassium ion. Therefore hydrogen
ions are preferentially discharged (reduced) to form hydrogen gas.
180
D
In dilute copper(II) chloride solution, there are four kinds of ions:
Cation
Anion
From copper(II) chloride
Cu2+(aq)
Cl-(aq)
From water
H+(aq)
OH-(aq)
The chloride ions and hydroxide ions are attracted to the anode. The copper(II) ions
and hydrogen ions are attracted to the cathode.
Option A –At the anode
The concentration of chloride ions in the solution is much greater than that
of hydroxide ions. Therefore chloride ions are preferentially discharged
(oxidized) to form chlorine gas.
Option B –The blue colour of the solution fades out because the concentration of copper(II)
ions decreases.
Option C –At the cathode
A copper(II) ion is a stronger oxidizing agent than a hydrogen ion. Hence,
copper(II) ions are preferentially discharged (reduced) to form a deposit of
copper on the cathode.
Option D –Copper(II) ions and chloride ions are consumed in the electrolysis. Thus the
solution becomes more dilute after some time.
181
A
182
A
183
A
During the electrolysis, the nickel is gradually transferred from the anode to the cathode.
The electrolyte remains the same.
In an electroplating process, the object to be plated (i.e. the iron spoon) is made the
cathode. The plating metal (i.e. the copper rod) is made the anode. An aqueous solution of
a salt of the plating metal is used as the electrolyte. The plating metal is gradually
transferred onto the object.
184
C
Option A –In the experiment, the nickel is gradually transferred onto the iron ring.
Therefore the mass of the iron ring increases.
Option C –The nickel(II) ions are reduced at the cathode.
Option D –The net effect is the transfer of nickel from the nickel electrode to the iron ring.
The rate at which nickel deposits on the iron ring is equal to the rate at which
the nickel electrode dissolves.
Therefore the concentration of nickel(II) ions in the solution remains the
same.
185
A
The tendency of sodium to lose electrons is higher than that of sulphur. Therefore sodium
atoms lose electrons and form sodium ions. Sulphur atoms gain these electrons and form
sulphide ions.
186
B
187
B
(2)
Electrons flow from electrode A to electrode B in the external circuit.
Reaction
SO2 + 2Mg  2MgO + S
Mg + 2AgNO3  Mg(NO3)2 + 2Ag
Pb3O4 + H2  3Pb + 4H2O
188
C
189
B
190
D
191
A
Explanation
SO2 is reduced.
Mg is oxidized.
Pb3O4 is reduced.
(1)
(2)
The service life of an alkaline manganese cell is relatively long.
Button shape silver oxide cells are suitable for quartz watches.
(1)
Zinc forms ions more readily than silver. Zinc atoms lose electrons and form zinc
ions. These ions go into the zinc nitrate solution.
Zn(s)  Zn2+(aq) + 2e(2)
(3)
192
B
(1)
Chlorine water oxidizes bromide ions to bromine (which is brown in colour).
Cl2(aq) + 2e-  2Cl-(aq)
2Br-(aq)  Br2(aq) + 2e-
(2)
Chlorine water oxidizes sulphite ions to sulphate ions (which is colourless).
Cl2(aq) + 2e-  2Cl-(aq)
SO32-(aq) + H2O(l)  SO42-(aq) + 2H+(aq) + 2eChlorine water oxidizes iodide ions to iodine (which is brown in colour).
(3)
193
B
Electrons flow from the zinc electrode to the silver electrode in the external circuit.
Silver ions near to the silver electrode gain electrons and form silver atoms.
Reduction occurs at the silver electrode.
Ag+(aq) + e-  Ag(s)
The salt bridge completes the circuit by allowing ions to move towards one halfcell from the other.
(2)
Cl2(aq) + 2e-  2Cl-(aq)
2I-(aq)  I2(aq) + 2eAcidified potassium dichromate solution can oxidize sodium sulphite solution, but
not sodium sulphate solution. The acidified potassium dichromate solution changes
from orange to green.
Cr2O72-(aq) + 14H+(aq) + 6e-  2Cr3+(aq) + 7H2O(l)
orange
green
194
C
0
–1
+1
Cl2(g) + 2NaOH(aq)  NaCl(aq) + NaOCl(aq) + H2O(l)
195
D
(1)
(2)
(3)
196
A
(1)
Iron(II) sulphate solution gives a green precipitate with dilute sodium hydroxide
solution while iron(III) sulphate solution gives a reddish brown precipitate.
Fe2+(aq) + 2OH-(aq)  Fe(OH)2(s)
green precipitate
Fe3+(aq) + 3OH-(aq)  Fe(OH)3(s)
reddish brown precipitate
Iron(II) sulphate solution decolourizes acidified potassium permanganate solution
while iron(III) sulphate solution does not.
MnO4-(aq) + 8H+(aq) + 5e-  Mn2+(aq) + 4H2O(l)
Fe2+(aq)  Fe3+(aq) + eIron(II) sulphate solution is pale green in colour while iron(III) sulphate
solution is yellow in colour.
An atom of X has an electronic arrangement of 2,7. X is fluorine. It is a strong
(3)
197
D
(1)
(2)
(3)
oxidizing agent.
X has an oxidation number of – in its compounds.
In the reaction between sodium sulphite solution and acidified potassium
permanganate solution, permanganate ions are reduced to manganese(II) ions.
MnO4-(aq) + 8H+(aq) + 5e-  Mn2+(aq) + 4H2O(l)
purple
pale pink
(or colourless)
In the reaction between sodium sulphite solution and acidified potassium
dichromate solution, dichromate ions are reduced to chromium(III) ions.
Cr2O72-(aq) + 14H+(aq) + 6e- 2Cr3+(aq) + 7H2O(l)
orange
green
In the reaction between sodium sulphite solution and bromine water, bromine is
reduced to bromide ions.
Br2(aq) + 2e- 2Br-(aq)
orange
colourless
(or brown)
198
B
(3)
Concentrated nitric acid is stored in brown bottles because it tends to decompose to
brown nitrogen dioxide gas.
199
A
(1)
In the left beaker, the purple colour of the solution fades out gradually. This is
because the permanganate ions MnO4-(aq) gain electrons and change to
manganese(II) ions Mn2+ (aq). The permanganate ions are reduced.
MnO4-(aq) + 8H+(aq) + 5e- Mn2+(aq) + 4H2O(l)
(2)
In the right beaker, a brown colour appears around the electrode. This is because
iodide ions I-(aq) lose electrons and change to iodine I2(aq). Iodide ions are
oxidized.
2I-(aq)  I2(aq) + 2eReduction occurs at electrode X.
(3)
200
A
(1)
In the left beaker, iron(II) ions Fe2+ (aq) lose electrons and change to iron(III) ions
Fe3+ (aq).
Fe2+(aq)  Fe3+(aq) + eIn the right beaker, dichromate ions Cr2O72-(aq) gain electrons and change to
chromium(III) ions.
Cr2O72-(aq) + 14H+(aq) + 6e- 2Cr3+(aq) + 7H2O(l)
Electrons flow from electrode X to electrode Y in the external circuit.
201
D
Concentrated nitric acid oxidizes most metals. The nitrate ion is reduced to nitrogen
dioxide, a brown gas.
NO3-(aq) + 2H+(aq) + e- NO2(g) + H2O(l)
202
A
203
B
204
A
(1)
(2)
(3)
Dilute hydrochloric acid reacts with iron powder only.
Dilute nitric acid reacts with both copper powder and iron powder.
Concentrated nitric acid reacts with both copper powder and iron powder.
(1)
Iron reacts with dilute nitric acid to give nitrogen monoxide gas. The gas gives a
(2)
(3)
brown gas when mixed with air. Iron reacts with dilute sulphuric acid to give
hydrogen gas.
Lead(II) nitrate solution gives a white precipitate (lead(II) sulphate) with dilute
sulphuric acid, but not dilute nitric acid.
Pb2+(aq) + SO42-(aq)  PbSO4(s)
Solid potassium carbonate reacts with both acids to give a colourless solution and
carbon dioxide gas.
CO32-(aq) + 2H+(aq)  H2O(l) + CO2(g)
Solid potassium carbonate dissolves in water to give mobile CO32–aq) ions.
205
D
(1)
Magnesium reacts with steam to give magnesium oxide and hydrogen.
Mg(s) + H2O(g)  MgO(s) + H2(g)
206
B
(1)
When dilute sulphuric acid reacts with calcium, insoluble calcium sulphate forms.
The calcium sulphate covers the surface of calcium and prevents further reaction.
Magnesium and concentrated nitric acid react to give nitrogen dioxide, not nitrogen
monoxide.
Mg(s) + 2NO3-(aq) + 4H+(aq)  Mg2+(aq) + 2NO2(g) + 2H2O(l)
(3)
207
C
(1)
Calcium reacts with water to give calcium hydroxide and hydrogen.
Ca(s) + 2H2O(l)  Ca(OH)2(s) + H2(g)
(3)
Zinc nitrate solution gives a white precipitate with dilute sodium hydroxide
solution. The precipitate dissolves in excess dilute sodium hydroxide solution.
Zn2+(aq) + 2OH-(aq)  Zn(OH)2(s)
208
C
(2)
(3)
209
D
Copper is a stronger reducing agent than hydroxide ions and sulphate ions.
Therefore the copper anode dissolves to form copper(II) ions (oxidized).
The net effect is the transfer of copper from the anode to the cathode. The rate at
which copper deposits on the cathode is equal to the rate at which the copper anode
dissolves. The concentration of copper(II) ions in the solution remains the same.
(1) & (2)
Electrolyte
Product at cathode
Concentrate sodium chloride solution Hydrogen
Dilute sodium chloride solution
hydrogen
(3)
210
A hydrogen ion is a stronger oxidizing agent than a sodium ion. Therefore
hydrogen ions are preferentially discharged at the cathode in both cases.
Zinc reacts with steam to give zinc oxide and hydrogen.
Zn(s) + H2O(g)  ZnO(s) + H2(g)
B
Electrolyte
Dilute hydrochloric acid
Dilute copper(II) chloride solution
Dilute sodium nitrate solution
211
D
212
B
Products of electrolysis
Hydrogen gas and chlorine gas
Copper and chlorine gas
Hydrogen gas and oxygen gas
During an electroplating process, the object to be plated (i.e. the iron fork) is made the
cathode. The plating metal (i.e. the copper rod) is made the anode. An aqueous solution of
a salt of the plating metal (i.e. a solution containing Cu2+(aq)) is used as the electrolyte.
The plating metal is gradually transferred onto the object.
(1)
(3)
The iron fork is the cathode.
The electrolyte contains Cu2+(aq).
213
D
214
C
Alkaline manganese cells are not rechargeable.
215
D
A zinc-carbon cell contains manganese(IV) oxide.
216
D
Dilute sulphuric acid is usually used to acidify potassium permanganate solution. Dilute
nitric acid is not used because it is an oxidizing agent and will react with reducing agents.
217
A
218
A
Metals above hydrogen in the electrochemical series react with dilute hydrochloric acid to
form hydrogen. These metals are stronger reducing agents than hydrogen. Therefore, they
reduce the hydrogen ions in the acid to hydrogen. The hydrogen ions oxidize the metals to
metal ions.
219
D
In an electrolytic cell, oxidation always occurs at the anode. Hydrogen ions are usually
preferentially discharged at the cathode.
220
A
221
C
In dilute potassium sulphate solution, there are four kinds of ions:
From potassium sulphate
Cation
Anion
K+(aq)
SO42-(aq)
From water
H+(aq)
OH-(aq)
The potassium ions and hydrogen ions are attracted to the cathode.
At the cathode
A hydrogen ion is a stronger oxidizing agent than a potassium ion. Therefore hydrogen
ions are preferentially discharged (reduced) to form hydrogen gas.
Section D
Short Questions
1
a)
b)
c)
d)
chemical
electrical
electrolyte
magnesium
e)
f)
g)
h)
i)
copper
copper
magnesium
magnesium
copper
2
Type of cell
Zinc-carbon cell
Materials of
electrodes
+ :Carbon
-: Zinc
Material of
electrolyte
Rechargeable?
(Yes/No)
Maximum voltage
Voltage over
discharge
Able to supply a
steady current?
(Yes/No)
Able to supply a
large current?
(Yes/No)
An example of
usage
Silver oxide cell
Ammonium
chloride
No
Alkaline
manganese cell
+ :Manganese(IV)
oxide
- :Zinc
Potassium
hydroxide
No
Potassium
hydroxide
No
Nickel-cadmium
cell
+ : Nickel (IV)
oxide
- : cadmium
Potassium
hydroxide
Yes
1.5 V
Falls rapidly
1.5 V
Falls slowly
No
Yes
1.5 V
Constant voltage
over discharge
Yes
1.25 V
Constant voltage
over discharge
Yes
No
Yes
No
Yes
Torches, small
radios, calculators,
remote control
units
(Any one)
Cassette players,
Discmans,
motorized toys,
flash guns, other
appliances where
there is heacy
+: Silver oxide
- : Zinc
Quartz watches,
calculators,
hearing aids,
pacemakers,
cameras, shavers
(Any one)
Electric shavers,
electric
toothbrushes,
motorized toys,
portable
telephones,
continuous use
(Any one)
portable tape
recorders, other
heavy duty motordriven appliances
(Any one)
3
Compound/ion
KMnO4
Na2SO4
FeCl2
H2S2O7
NH4NO3
Cr2O72NH4VO3
[Pb(OH)4]2-
Oxidation number of the underlined elements
+7
+6
+2
+6
-3
+6
+5
+2
4
Equation
+2
-2 +1 +2 –2 +1
2+
a) Cu + 2OH-  Cu(OH)3
+3
0
+5 -1
b) PCl3 + Cl2  PCl5
A redox reaction?
No
+4
0
+2
0
c) SO2 + 2Mg  2MgO + S
Yes
Yes
+1+1+4–2 +1 +4–2 +1–2 +4-2 No
d) 2NaHCO3  Na2CO3 + H2O + CO2
0
+1
+2
0
Yes
e) Zn + 2AgNO3  Zn(NO3)2 + 2Ag
-3
+2
0
0
Yes
f) 2NH3 + 3CuO  3Cu + N2 + 3H2O
+6–2
+1
+6 –2 +1 -2
2+
g) 2CrO4 + 2H  Cr2O72- + H2O
+5–2
-1
0
h) 2KClO3  2KCl + 3O2
No
Yes
Yes
i) Pb3O4 + 4H2  3Pb + 4H2O
+2 +5 –2
+1 +6–2
+2+6 –2 +1+5 –2
j) Pb(NO3)2 + H2SO4  PbSO4 2HNO3
No
Explanation
The oxidation numbers of all the atoms
remain unchanged.
The oxidation number of P changes from +3
to +5 while that of Cl changes from 0 to -1.
PCl3 is oxidized and Cl2 is reduce.
The oxidation number of S changes from +4
to 0 while that of Mg changes from 0 to +2.
SO2 is reduced and Mg is oxidized.
The oxidation numbers of all the atoms
remain unchanged.
The oxidation number of Zn changes from 0
to +2 while that of Ag changes from +1 to
0.Zn is oxidized and AgNO3 is reduced.
The oxidation number of N changes from -3
to 0 while that of Cu changes from +2 to 0.
NH3 is oxidized and CuO is reduced.
The oxidation numbers of all the atoms
remain unchanged.
The oxidation number of Cl changes from
+5 to -1 while that of O changes from -2 to
0.
Pb3O4 loses oxygen while H2 gain oxygen.
Pb3O4 is reduced and H2 is oxidized.
The oxidation numbers of all the atoms
remain unchanged.
5
Equation
+2
0
a) Mg + CuSO4  MgSO4 + Cu
+3
+3
b) Fe3+ + 3OH-  Fe(OH)3
Is the underlined
reactant reduced?
Yes
No
No
c) CaCO3 + 2HCl  CaCl2 + CO2 + H2O
+1
0
d) 2Al + 6HCl  2AlCl3 + 3H2
Yes
+3
+2
e) Fe2(SO4)3 + H2S  2FeSO4 + S+ H2SO4
Yes
-1
0
f) 2KBr + Cl2  2KCl + Br2
No
6
a)
electricity
b)
c)
d)
e)
f)
g)
h)
i)
chemical reactions
negative
positive
cathode
negative
positive
anode
cathode
Explanation
The oxidation number of Cu changes
from +2 to 0. Therefore CuSO4 is
reduced/
The oxidation number of Fe remains
unchanged. Therefore Fe3+ is not
reduced.
The reaction is not a redox reaction.
The oxidation numbers of all the
atoms remain unchanged.
The oxidation number of H changes
form +1 to 0. Therefore HCl is
reduced.
The oxidation number of Fe changes
from +3 to +2. Therefore Fe2(SO4)3 is
reduced.
The oxidation number of Br changes
from -1 to 0. Therefore KBr is
oxidized, not reduced.
7
Solution
Dilute sulphuric
acid
Dilute
hydrochloric
acid
Very dilute
sodium chloride
Dilute sodium
chloride
Dilute sodium
sulphate
Concentrated
sodium chloride
Dilute
copper(II)
sulphate
Dilute
copper(II)
sulphate
Dilute
copper(II)
sulphate
Dilute
copper(II)
chloride
Materials of
anode
cathode
Platinum
Platinum
Products at
anode
cathode
Oxygen
Hydrogen
Carbon
Chlorine
Hydrogen
Becomes more dilute
Carbon
Carbon
Oxygen
Hydrogen
Becomes more concentrated
Carbon
Carbon
Chlorine
Hydrogen
Carbon
Carbon
Oxygen
Hydrogen
Becomes sodium hydroxide
solution
Becomes more concentrated
Carbon
Mecury
Chlorine
Sodium
Becomes more dilute
Carbon
Carbon
Oxygen
Copper
Becomes sulphuric acid solution
Carbon
Copper
Oxygen
Copper
Becomes sulphuric acid solution
Copper
Copper
Copper(II)
ions
Copper
Remains the same
Carbon
Carbon
Chlorine
Copper
Becomes more dilute
Structured Questions
1
i)
Manganese(IV) oxide
ii)
Zinc
Potassium hydroxide
 Voltage falls slowly over discharge
b)
c)
d)
e)
Becomes more concentrated
Carbon
Section E
a)
Change in the solution
 Able to supply a steady current
 Able to supply a large current
 Long shelf life
 Long service life
 Leak proof
More expensive
 Discmans
 Motorized toys
 Flash guns
 Appliances where there are heavy continuous use
Any two
Any one
2
a)
i)
The size of an alkaline manganese cell is too large.
ii)
b)
3
a)
b)
A toy car with an electric motor requires a large current. A zinc-carbon cell is used
when small currents are needed.
iii)
The size of a silver oxide cell is small and cannot supply enough electrical energy
for torches.
i)
A:
Zn + 2OH– ZnO + H2O + 2e–
B:
Ag2O + H2O + 2e– 2Ag + 2OH–
ii)
Electrode B is the positive electrode. This is because reduction occurs at this
electrode.
i)
Sodium and gold
ii)
No.
 Sodium reacts vigorously with the electrolyte.
 Gold is too expensive.
The metals are cheap.
i)
 The voltage of the cell is small.
Any one
 Iron will rust.
Zinc and copper (students may choose other metals).
Advantages:
 The metals are relatively cheap.
 The metals are far apart in the electrochemical series. Therefore the voltage of the cell is
ii)
c)
relatively high.
 The metals do not react with water and thus are safe.
4
a)
There is no right or wrong answer to this question. Choices may be made according to one
of the following criteria:
 Alkaline manganese cell may be used in every case for the sake of convenience. This
saves the trouble of replacing dry cells frequently and recharging of cells.
 Choices may be made based on the ‘cost-effectiveness’of the different types of cell.
Type
Operating cost
Flash gun
Walkman
Radio
(HK$ per cycle)
(HK$ per hour)
(HK$ per hour)
Zinc-carbon cell
0.08
0.3
0.024
Alkaline manganese cell 0.02
0.29
0.029
b)
5
a)
It is more cost effective to use alkaline manganese cell for flash gun and
walkman, while it is more cost effective to use zinc-carbon cell for radio.
 Rechargeable nickel-cadmium cell may be used if dry cells are used for a long time
every day.
Cd + 2OH-  Cd(OH)2 + 2eNiO2 + 2H2O + 2e-  Ni(OH)2 + 2OHHydrogen
b)
Z, Y, zinc, X
Z is the most reactive because only Z reacts with the water in zinc sulphate solution to give
hydrogen. Y is more reactive than zinc because it displaces zinc from zinc sulphate
solution. X is less reactive than zinc because it cannot displace zinc from zinc sulphate
solution. Therefore X is the least reactive.
6
i)
a)
ii)
Zn2+(aq) + 2e-  Zn(s)
Mg(s)  Mg2+(aq) + 2eX2+(aq) + 2e-  X(s)
b)
i)
ii)
7
 Y is suitable because it is higher than zinc in the reactivity series. Therefore it
forms ions more readily than zinc. This matches with the direction of electron
flow in the external circuit of the chemical cell.
 Y does not react with water.
c)
Oxidation occurs at the magnesium strip. This is because magnesium atoms lose
electrons to form magnesium ions.
Reduction occurs at the strip made of X. This is because X2+ ions gain electrons to
form atoms.
a)
As a salt bridge.
b)
 Digital multimeter
 Voltmeter
Any one
c)
d)
Y, Ni, X, Z
In the cell formed by connecting half-cells A and C, electrons flow from nickel to Y in the
external circuit. This is because nickel forms ions more readily than Y.
In the cell formed by connecting half-cells A and B, electrons flow from X to
nickel in the external circuit. This is because X forms ions more readily than nickel.
In the cell formed by connecting half-cells B and D, electrons flow from Z to X in
the external circuit. This is because Z forms ions more readily than X.
To summarize, Z forms ions most readily while Y forms ions least readily.
8
e)
No change would be observed. This is because X forms ions more readily than nickel and
thus X is more reactive than nickel. No displacement reaction would take place.
a)
The brown iodine solution becomes colourless. This is because brown iodine is reduced to
colourless iodide ions.
I2(aq) + SO32-(aq) + H2O(l)  2I-(aq) + SO42-(aq) + 2H+(aq)
b)
The purple colour of acidified potassium permanganate solution fades out. This is because
the purple permanganate ions are reduced to pale pink / colourless manganese(II) ions.
2MnO4-(aq) + 10Cl-(aq) + 16H+(aq)  2Mn2+(aq) + 5Cl2(aq) + 8H2O(l)
c)
A brown colour develops. This is due to the formation of iodine in the reaction.
Cl2(aq) + 2I-(aq)  2Cl-(aq) + I2(aq)
d)
The orange acidified potassium dichromate solution turns green. This is because the
orange dichromate ions are reduced to green chromium(III) ions.
Cr2O72-(aq) + 3SO32-(aq) + 8H+(aq)  2Cr3+(aq) + 3SO42-(aq) + 4H2O(l)
e)
The purple acidified potassium permanganate solution turns greenish yellow. This is
because iron(II) ions are oxidized to yellow iron(III) ions and there is excess pale green
iron(II) ions in the reaction mixture.
MnO4-(aq) + 5Fe2+(aq) + 8H+(aq)  Mn2+(aq) + 5Fe3+(aq) + 4H2O(l)
f)
A brown gas is given off. This is because nitrate ions are reduced to brown nitrogen
dioxide gas.
SO32-(aq) + 2NO3- (aq) + 2H+ (aq) SO42-(aq) + 2NO2(g) + H2O(l)
g)
The zinc granules dissolve and a brown gas is given off. This is because nitrate ions are
reduced to brown nitrogen dioxide gas.
Zn(s) + 2NO3-(aq) + 4H+(aq)  Zn2+(aq) + 2NO2(g) + 2H2O(l)
h)
The copper dissolves. A colourless gas is given off. This gas gives a brown gas when
mixed with air. A blue solution forms. This is because nitrate ions are reduced to
colourless nitrogen monoxide gas. This gas gives brown nitrogen dioxide gas when mixed
with air. The resulting solution contains copper(II) ions and thus blue in colour.
3Cu(s) + 2NO3-(aq) + 8H+(aq)  3Cu2+(aq) + 2NO(g) + 4H2O(l)
9
10
a)
i)
ii)
iii)
Sulphite ions reduce the orange bromine to colourless bromide ions.
Sodium sulphite solution acts as a reducing agent.
SO32-(aq) + Br2(aq) + H2O(l)  SO42-(aq) + 2Br-(aq) + 2H+(aq)
b)
i)
ii)
iii)
Chlorine water oxidizes the colourless bromide ions to orange bromine.
Chlorine water acts as an oxidizing agent.
Cl2(aq) + 2Br-(aq)  2Cl-(aq) + Br2(aq)
a)
Cr2O3(s) + 2Al(s)  2Cr(s) + Al2O3(s)
b)
i)
Aluminium is oxidized. This is because the oxidation number of aluminium
increases from 0 to +3.
ii)
Chromium(III) oxide is reduced. This is because the oxidation number of
chromium decreases from +3 to 0.
c)
Pure chromium is very expensive.
d)
Cr2O3(s) + 2Al(s)  2Cr(s) + Al2O3(s)
500 g
?g
Number of moles of Cr2O3
=
=
Mass
Molar mass
500 g
152.0 g mol -1
= 3.29 mol
According to the equation, 1 mole of Cr2O3 reacts with 2 moles of Al to give 2 moles of
Cr.
∴
Number of moles of Cr obtained = 2 x 3.29 mol = 6.58 mol
Mass of Cr obtained = Number of moles of Cr x Molar mass of Cr
= 6.58 mol x 52.0 g mol-1
= 342 g
11
a)
Fe2+(aq)  Fe3+(aq) + e-
Iron(II) ions are oxidized. This is because the oxidation number of iron increases from +2
to +3.
12
13
b)
Cr2O72-(aq) + 14H+(aq) + 6e-  2Cr3+(aq) + 7H2O(l)
Dichromate ions are reduced. This is because the oxidation number of chromium decreases
from +6 to +3.
c)
i)
ii)
d)
Electrons flow from beaker X to beaker Y in the external circuit.
e)
Beaker X –The iron(II) sulphate solution changes from pale green to yellow.
Beaker Y –The acidified potassium dichromate solution changes from orange to green.
f)
Cr2O72-(aq) + 6Fe2+(aq) + 14H+(aq)  2Cr3+(aq) + 6Fe3+(aq) + 7H2O(l)
a)
The colour change occurs because yellow iron(III) ions gain electrons and change
to green iron(II) ions.
Fe3+(aq) + e-  Fe2+(aq)
b)
c)
A brown colour appears around the carbon electrode. This is because colourless iodide
ions lose electrons and change to iodine.
2I-(aq)  I2(aq) + 2eFrom left to right
d)
2Fe3+(aq) + 2I-(aq)  2Fe2+(aq) + I2(aq)
e)
i)
Acidified potassium dichromate solution is the oxidizing agent.
Iron(II) sulphate solution is the reducing agent.
ii)
To complete the circuit by allowing ions to move towards one half-cell from the
other.
No. The sodium sulphite solution will react with the iron(III) sulphate solution.
a)
i)
ii)
Carbon rod
2NH4+(aq) + 2e-  2NH3(aq) + H2(g)
b)
i)
Zinc case
ii)
Zn(s)  Zn2+(aq) + 2eHydrogen is produced and collected on the surface of the positive electrode. Since
hydrogen is a poor conductor of electricity, the accumulation of hydrogen at the positive
electrode may hinder further reactions and decrease the current of the cell. Manganese(IV)
c)
oxide, an oxidizing agent, is used to remove the hydrogen.
d)
If a current is drawn from the cell rapidly, the gaseous product cannot be removed fast
enough. The voltage drops as a result.
e)
The zinc case undergoes oxidation to give zinc ions in the cell reaction.
f)
The materials inside the cells do not decompose even after a long time. These materials
may combine with other compounds and form harmful substances which pollute the
environment.
g)
There is a slow direct reaction between the zinc electrode and ammonium ions. After some
time, the zinc case becomes too thin and the paste leaks out. This may cause damage to
electric appliances.
14
15
a)
i)
ii)
Corrosive property
Oxidizing hazard warning symbol
b)
i)
ii)
iii)
Copper
Cu(s) + 2NO3-(aq) + 4H+(aq)  Cu2+(aq) + 2NO2(g) + 2H2O(l)
By filtration
iv)
v)
Copper(II) nitrate solution
Cu2+(aq) + 2OH-(aq)  Cu(OH)2(s)
a)
X is dilute nitric acid.
Y is dilute sulphuric acid.
X (dilute nitric acid) reacts with calcium to give colourless nitrogen monoxide gas.
When this gas mixes with air, it reacts with oxygen in the air to give brown nitrogen
dioxide gas.
Y (dilute sulphuric acid) reacts with calcium to give colourless hydrogen gas.
Insoluble calcium sulphate is produced also. The calcium sulphate covers the surface of
calcium. This prevents further reaction between calcium and the acid. Therefore the
reaction between calcium and Y soon slows down and then stops.
b)
i)
ii)
Ca(s) + 2H+(aq)  Ca2+(aq) + H2(g)
3Ca(s) + 2NO3-(aq) + 8H+(aq)  3Ca2+(aq) + 2NO(g) + 4H2O(l)
16
a)
Sugar solution is a non-conductor of electricity.
b)
i)
ii)
c)
X -oxygen
Y -hydrogen
At electrode X
4OH-(aq)  O2(g) + 2H2O(l) + 4eAt electrode Y
2H+(aq) + 2e-  H2(g)
i)
The concentration of iodide ions in the solution is much greater than that of
hydroxide ions. Iodide ions are preferentially discharged to form iodine.
ii)
A hydrogen ion is a stronger oxidizing agent than a sodium ion. Therefore
hydrogen ions are preferentially discharged to form hydrogen gas.
Hydrogen ions and iodide ions are consumed in the electrolysis. Sodium ions and
hydroxide ions remain in the solution. Eventually, the solution becomes sodium
iii)
hydroxide solution.
d)
17
18
i)
At electrode X
4OH-(aq)  O2(g) + 2H2O(l) + 4eAt electrode Y
Ag+(aq) + e-  Ag(s)
ii)
Silver ions and hydroxide ions are consumed in the electrolysis. Hydrogen ions and
nitrate ions remain in the solution. The solution eventually becomes nitric acid
solution.
a)
Calcium hydroxide
b)
i)
ii)
c)
It is because molten magnesium chloride contains mobile ions.
a)
Oxygen
b)
The gas relights a glowing splint.
c)
i)
Reaction with dilute hydrochloric acid
Mg(OH)2(s) + 2HCl(aq)  MgCl2(aq) + 2H2O(l)
Use concentrated sodium chloride solution instead of very dilute sodium chloride
solution. The concentration of chloride ions in the solution is much greater than
that of hydroxide ions. Therefore chloride ions are preferentially discharged to
ii)
iii)
19
a)
form chlorine gas.
Sodium hydroxide solution
 Manufacture of bleach
 Manufacture of soaps and detergents
Any one
At carbon electrode X
The sulphate ions and hydroxide ions are attracted to electrode X. A hydroxide ion is a
stronger reducing agent than a sulphate ion. Therefore hydroxide ions are preferentially
discharged.
At carbon electrode Y
The copper(II) ions and hydrogen ions are attracted to the electrode Y. A copper(II) ion is
a stronger oxidizing agent than a hydrogen ion. Therefore copper(II) ions are preferentially
discharged to form a deposit of copper on electrode Y.
b)
At carbon electrode X
4OH- (aq)  O2(g) + 2H2O(l) + 4eAt carbon electrode Y
Cu2+(aq) + 2e-  Cu(s)
c)
The solution becomes sulphuric acid. This is because copper(II) ions and hydroxide ions
are consumed in the electrolysis. Hydrogen ions and sulphate ions remain in the solution.
20
d)
The net effect is the transfer of copper from electrode X to electrode Y. The dilute
copper(II) sulphate solution remains the same.
a)
i)
ii)
At the anode
2Cl-(aq)  Cl2(g) + 2eAt the cathode
Na+(aq) + e- + Hg(l)  Na/Hg(l)
b)
i)
ii)
Sodium hydroxide and hydrogen
The sodium formed dissolves in the mercury cathode to form an alloy called
sodium amalgam. The sodium amalgam then moves towards the water. The sodium
reacts with water to form sodium hydroxide and hydrogen.
2Na/Hg(l) + 2H2O(l)  2NaOH(aq) + H2(g) + 2Hg(l)
c)
Sodium ions and chloride ions are consumed in the electrolysis. Thus the sodium chloride
solution becomes more and more dilute.
21
a)
i)
The liquid around pencil A turns red.
ii)
Around pencil A, hydroxide ions are preferentially discharged to form oxygen gas.
4OH-(aq)  O2(g) + 2H2O(l) + 4eWater ionizes continuously to replace the hydroxide ions discharged. Thus
there is an excess of hydrogen ions near pencil A and the solution there becomes
acidic. The solution turns the universal indicator red.
b)
i)
The liquid around pencil B turns blue.
ii)
Around pencil B, hydrogen ions are preferentially discharged to form hydrogen gas.
2H+(aq) + 2e-  H2(g)
Water ionizes continuously to replace the hydrogen ions discharged. Thus
there is an excess of hydroxide ions near pencil B. The solution there becomes
alkaline. The solution turns the universal indicator blue.
22
a)
i)
The chloride ions and hydroxide ions in the copper(II) chloride solution move
ii)
towards the anode in the electrolysis process. The concentration of chloride ions in
the solution is much greater than that of hydroxide ions. Therefore chloride ions are
preferentially discharged to form chlorine gas.
Cu2+(aq) + 2e-  Cu(s)
b)
It is because the concentration of copper(II) ions in the solution decreases.
c)
i)
ii)
Oxygen is given off at the anode.
Hydrogen is given off at the cathode.
4OH-aq)  O2(g) + 2H2O(l) + 4e2H+(aq) + 2e-  H2(g)
23
a)
i)
ii)
Copper electrode
At magnesium electrode
Mg(s)  Mg2+(aq) + 2eAt copper electrode
Cu2+(aq) + 2e-  Cu(s)
b)
i)
Electrode Y
ii)
At electrode X
4OH-(aq)  O2(g) + 2H2O(l) + 4eAt electrode Y
2H+(aq) + 2e-  H2(g)
24
c)
When hydrogen ions and hydroxide ions are discharged, more water molecules ionize. The
net effect is that water is decomposed. The number of hydrogen ions and sulphate ions
from the sulphuric acid remains the same. The concentration of sulphuric acid increases as
water is consumed in the electrolysis.
a)
i)
Reaction with dilute sulphuric acid
ii)
CuO(s) + H2SO4(aq)  CuSO4(aq) + H2O(l)
i)
ii)
Add copper to displace ions of less reactive metals.
In the electrochemical series, positions of metals that are more reactive than copper
b)
are higher than that of copper. Copper(II) ions are preferentially discharged in the
electrolysis.
c)
i)
ii)
At the anode
4OH-(aq)  O2(g) + 2H2O(l) + 4eAt the cathode
Cu2+(aq) + 2e-  Cu(s)
25
a)
Electroplating is the coating of an object with a thin layer of a metal by electrolysis.
b)
To make the knob conduct electricity.
c)
The solution contains mobile ions.
d)
The nickel electrode
e)
Ni2+(aq) + 2e-  Ni(s)
f)
The concentration of nickel(II) ions in the electrolyte can be maintained.
g)
 To recover the nickel metal.
 Nickel(II) ions are harmful to marine lives. Humans may get poisoned by eating such
things.
26
a)
The ammeter is an instrument used to measure the electric current passing through the
circuit.
The rheostat is used to vary the resistance in the circuit and regulate the current.
b)
In solid state, the ions in potassium iodide are held together by strong attraction. They are
not free to move. Therefore solid potassium iodide does not conduct electricity. When
water is added to the compound, the compound dissolves in the water and the ions
become mobile. A current can flow through the solution.
c)
i)
A brown colour develops around electrode A.
A colourless gas is given off from electrode B.
ii)
At electrode A
2I-(aq)  I2(aq) + 2eAt electrode B
2H+(aq) + 2e-  H2(g)
d)
i)
ii)
iii)
27
Electrode C dissolves.
A reddish brown deposit forms on electrode D.
At electrode C
Cu(s)  Cu2+(aq) + 2eAt electrode D
Cu2+(aq) + 2e-  Cu(s)
The blue colour of the dilute copper(II) sulphate solution does not change. It is
because the concentration of copper(II) ions in the solution remains the same.
[Remark: You may discuss redox reactions with reference to electron transfer or change in
oxidation number.]
Oxidation and reduction can be defined in terms of electron transfer / change in oxidation number.
Oxidation and reduction always occur at the same time. The combined process is called a redox
reaction.
When defined in terms of electron transfer, oxidation is the process in which a substance
loses electrons while reduction is the process in which a substance gains electrons. In the reaction
between magnesium and copper(II) sulphate solution, each magnesium atom loses two electrons
to form a magnesium ion. On the other hand, each copper(II) ion gains two electrons to form a
copper atom. Magnesium undergoes oxidation and copper undergoes reduction. Therefore the
reaction is a redox reaction.
When defined in terms of change in oxidation number, oxidation involves an increase in
oxidation number while reduction involves a decrease in oxidation number. In the reaction
between magnesium and copper(II) sulphate solution, the oxidation number of magnesium
increases from 0 to +2. On the other hand, the oxidation number of copper decreases from +2 to 0.
Magnesium undergoes oxidation and copper undergoes reduction. Therefore the reaction is a
redox reaction.
28
During electrolysis, anions give up electrons at the anode and are oxidized. Anions at higher
positions of the electrochemical series are more readily discharged because they are stronger
reducing agents (i.e. they lose electrons more readily). During electrolysis, cations gain electrons
at the cathode and are reduced. Cations at lower positions of the electrochemical series are more
readily discharged because they are stronger oxidizing agents (i.e. they gain electrons more
readily).
In dilute copper(II) sulphate solution, there are four kinds of ions: copper(II) ions,
hydrogen ions, sulphate ions and hydroxide ions. During the electrolysis of dilute copper(II)
sulphate solution using carbon electrodes, the sulphate ions and hydroxide ions move towards the
anode while the copper(II) ions and hydrogen ions move towards the cathode.
At the anode, hydroxide ions are oxidized to form oxygen gas. This is because a hydroxide
ion is a stronger reducing agent than a sulphate ion. Hence hydroxide ions are preferentially
discharged.
At the cathode, copper(II) ions are reduced to form a deposit of copper. This is because a
copper(II) ion is a stronger oxidizing agent than a hydrogen ion. Hence copper(II) ions are
preferentially discharged.
29
[Remark: You may either discuss the refining of copper or electroplating.]
Electrolysis is used industrially in various ways. One of these uses, refining of copper /
electroplating, will be discussed here.
During the refining of copper, the electrolytic cell contains a solution of copper(II)
sulphate and sulphuric acid as the electrolyte. The impure copper becomes the anode. The cathode
is a thin sheet of very pure copper. When the cell is operating at the correct voltage, iron and zinc
impurities in the anode give up electrons first. Then copper gives up electrons to form copper(II)
ions. Impurities such as silver, gold and platinum settle at the bottom of the container. Copper(II)
ions gain electrons at the cathode to form copper. Copper is gradually transferred from the anode
to the cathode during the process.
Electroplating is the coating of an object with a thin layer of a metal by electrolysis. This is
done to protect the object from corrosion or to improve its appearance. In this process, the object
to be plated is cleaned and made the cathode. The plating metal is made the anode. An aqueous
solution of a salt of the plating metal is used as the electrolyte. The plating metal is gradually
transferred onto the ob
ject. Objects may be electroplated with copper, nickel, chromium, gold or silver.
30
First dissolve the copper(II) sulphate crystals in the beaker of distilled water. Then connect the
positive terminal of the d.c. power supply to the copper plate and the negative terminal of the d.c.
power supply to the metal key. Dip both the copper plate and the key into the copper(II) sulphate
solution.
Pass electricity through the circuit for some time. The copper plate becomes thinner. A
layer of shiny metal deposits onto the key.
31
Acids and alkalis in effluents from electroplating factories cause changes in the pH value of water
and affect water life. Ions of heavy metals (such as nickel, chromium and mercury) are readily
absorbed into the bodies of living things such as shellfish and plants. Humans may get poisoned
by eating such things. They cause abnormal body functions.
To control the pH value of effluents, we can add slaked lime (calcium hydroxide) to the
waste to neutralize the acid. On the other hand, we can control the pH value of alkaline waste by
adding sulphuric acid to it.
Most heavy metal hydroxides are insoluble in water. We can add sodium hydroxide
solution to the effluents. This reacts with the heavy metal ions to form insoluble metal hydroxides.
The solid is then filtered off.